Question

A 300 kg merry-go-round in the shape of a horizontal disk with a radius of 1.9 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.6 rad/s in 3.5 s

Answer 1

557.7Nm

Step-by-step explanation:

We are given that

`m=300 kg`

Radius,r=1.9 m

Angular speed,
`\omega=3.6rad/s`

Time,t=3.5 s

Initial angular velocity,
`\omega_0=0`

We have to find the large torque would have to be exerted .

`\alpha=(\omega-\omega_0)/(t)=(3.6-0)/(3.5)=1.03rad/s^2`

`\tau=I\alpha`

`\tau=(1)/(2)mr^2\alpha`

Where
`I=(1)/(2)mr^2`

Substitute the values

`\tau=(1)/(2)(300)(1.9)^2(1.03)=557.7Nm`